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Calculus

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If the volume of a cube is increasing at 24 in^3/min and the surface area of the cube is increasing at 12 in^2/min, what is the length of each edge of the cube?

I know that dV/dt=24 and ds/ds=12. I also know that Volume=s^3 and Surface Area=6*s^2, but what do I do from there?

  • Calculus - ,

    from V = x^3
    dV/dt = 3x^2 dx/dt
    dx/dt = 24/(3x^2) = 8/x^2

    from A = 6x^2
    dA/dt = 12x dx/dt
    dx/dt = 12/(12x) = 1/x

    then 8/x^2 = 1/x
    x^2 = 8x
    x = 8

  • Calculus - ,

    Let the independent variable be
    s = length of one side of the cube.
    As you mentioned,
    V(s) = s³
    S(s) = 6s²

    dV/dt = dV/ds*ds/dt = 3s² ds/dt = 24 ... (1)
    dS/ds = dS/ds*ds/dt = 12s ds/dt = 12 ...(2)

    Solve for s and ds/dt by substitution or dividing (1) by (2).

  • Calculus - ,

    Let the independent variable be
    s = length of one side of the cube.
    As you mentioned,
    V(s) = s³
    S(s) = 6s²

    dV/dt = dV/ds*ds/dt = 3s² ds/dt = 24 ... (1)
    dS/ds = dS/ds*ds/dt = 12s ds/dt = 12 ...(2)

    Solve for s and ds/dt by substitution or dividing (1) by (2).

    I get s=8, and ds/dt=1/8.

  • Calculus - ,

    8

  • Calculus - ,

    5

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